$$2\cdot\dfrac{5-t}{5}t^2(t-5)=\dfrac{-2t^4+20t^3-50t^2}{5}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2 \cdot \frac{5-t}{5}t^2(t-5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-2t+10}{5}t^2(t-5) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-2t^3+10t^2}{5}(t-5) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-2t^4+20t^3-50t^2}{5}\end{aligned} $$ | |
| ① | Multiply $2$ by $ \dfrac{5-t}{5} $ to get $ \dfrac{-2t+10}{5} $. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{5-t}{5} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{5-t}{5} \xlongequal{\text{Step 2}} \frac{ 2 \cdot \left( 5-t \right) }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 10-2t }{ 5 } = \frac{-2t+10}{5} \end{aligned} $$ |
| ② | Multiply $ \dfrac{-2t+10}{5} $ by $ t^2 $ to get $ \dfrac{ -2t^3+10t^2 }{ 5 } $. Step 1: Write $ t^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{-2t+10}{5} \cdot t^2 & \xlongequal{\text{Step 1}} \frac{-2t+10}{5} \cdot \frac{t^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( -2t+10 \right) \cdot t^2 }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -2t^3+10t^2 }{ 5 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{-2t^3+10t^2}{5} $ by $ t-5 $ to get $ \dfrac{-2t^4+20t^3-50t^2}{5} $. Step 1: Write $ t-5 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{-2t^3+10t^2}{5} \cdot t-5 & \xlongequal{\text{Step 1}} \frac{-2t^3+10t^2}{5} \cdot \frac{t-5}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( -2t^3+10t^2 \right) \cdot \left( t-5 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -2t^4+10t^3+10t^3-50t^2 }{ 5 } = \frac{-2t^4+20t^3-50t^2}{5} \end{aligned} $$ |